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Math Help - equation of hyperbola

  1. #1
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    equation of hyperbola

    Hi! I need help to solve a task.

    Write an equation of hyperbola. Foci coincides with the ellipse (x^2/35+y^2/10=1) focal points. There is given also a point from hyperbola - L(5sqrt(2);4)

    Thanks for help in advance.
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  2. #2
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    Re: equation of hyperbola

    Quote Originally Posted by Kiiefers View Post
    Hi! I need help to solve a task.

    Write an equation of hyperbola. Foci coincides with the ellipse (x^2/35+y^2/10=1) focal points. There is given also a point from hyperbola - L(5sqrt(2);4)

    Thanks for help in advance.
    1. With an ellipse the foci have the coordinates F_1(-e,0)~,~F_2(e,0) with e = \sqrt{a^2-b^2}
    That means the foci are at F_1(-5,0) and F_2(5,0)

    2. A hyperbola is defined by a constant difference of the distances between a point L on the hyperbola and the foci:

    |LF_1|-|LF_2| = 2a

    with L(5\sqrt{2},4)

    I've got a = \frac12 \cdot \sqrt{182-2\sqrt{3281}}

    3. The excentricity of a hyperbola is calculated by:

    e = \sqrt{a^2+b^2}

    Since the excentricity is e = 5 you can determine the value of b.

    I've got b=\frac12 \cdot \sqrt{2\sqrt{3281}-82}

    (For this part I used my computer)

    4. Therefore the equation of the hyperbola is:

    \frac{x^2}{\left( \frac12 \cdot \sqrt{182-2\sqrt{3281}}   \right)^2}-\frac{y^2}{\left( \frac12 \cdot \sqrt{2\sqrt{3281}-82} \right)^2}=1

    5. I've attached a drawing of the ellipse and the hyperbola with the point L on the right branch.
    Attached Thumbnails Attached Thumbnails equation of hyperbola-hypundellips.png  
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  3. #3
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    Re: equation of hyperbola

    Thanks for help.
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